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Constant Current LED Circuit
Make adjustable mosfet constant current sourceSimple LED Dimming from Analogue Voltage ControlMOSFET not turning offProblem with arduino controlled constant current supply with MOSFET/NPNLED Matrix Constant Current Driver with MOSFET and NPN TransLED Intensity variation with Temperature (At constant Forward Current)?can leds take varying voltage > listed voltage, with constant current?PWM equivalent for a constant current source and load18v vs 12v battery questionConstant current source gets heated
$begingroup$
In the below circuit, the circuit maintains a constant current on LED.
My question is after the LED there is a voltage of 18V, still, need to be dropped. Will the NPN Transistor heat up while dropping the voltage?
Also, the voltage source to led will vary from 12V to 20V. How can we modify the circuit? Or the existing circuit is fine?
led npn
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
$endgroup$
|
show 8 more comments
$begingroup$
In the below circuit, the circuit maintains a constant current on LED.
My question is after the LED there is a voltage of 18V, still, need to be dropped. Will the NPN Transistor heat up while dropping the voltage?
Also, the voltage source to led will vary from 12V to 20V. How can we modify the circuit? Or the existing circuit is fine?
led npn
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
$endgroup$
2
$begingroup$
The polarity of your left-hand transistor is wrong. The transistor is operating in reverse mode (which it will do, just not very well). Try connecting the emitter to ground.
$endgroup$
– WhatRoughBeast
15 hours ago
$begingroup$
Right. What difference will it make by the way?
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
The transistor will have less gain in the reversed configuration than in the normal. So in normal configuration the current regulation will be greater for supply voltage and load variations.
$endgroup$
– WhatRoughBeast
15 hours ago
2
$begingroup$
It is good practice to number the components R1, R2, Q1, Q2, etc. so we can discuss them without confusion.
$endgroup$
– Transistor
15 hours ago
$begingroup$
@Transistor You are right, but the tool didn't allow me to add the component name. Apologies.
$endgroup$
– Embedded Geek
15 hours ago
|
show 8 more comments
$begingroup$
In the below circuit, the circuit maintains a constant current on LED.
My question is after the LED there is a voltage of 18V, still, need to be dropped. Will the NPN Transistor heat up while dropping the voltage?
Also, the voltage source to led will vary from 12V to 20V. How can we modify the circuit? Or the existing circuit is fine?
led npn
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
$endgroup$
In the below circuit, the circuit maintains a constant current on LED.
My question is after the LED there is a voltage of 18V, still, need to be dropped. Will the NPN Transistor heat up while dropping the voltage?
Also, the voltage source to led will vary from 12V to 20V. How can we modify the circuit? Or the existing circuit is fine?
led npn
led npn
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
asked 15 hours ago
Embedded GeekEmbedded Geek
16913
16913
2
$begingroup$
The polarity of your left-hand transistor is wrong. The transistor is operating in reverse mode (which it will do, just not very well). Try connecting the emitter to ground.
$endgroup$
– WhatRoughBeast
15 hours ago
$begingroup$
Right. What difference will it make by the way?
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
The transistor will have less gain in the reversed configuration than in the normal. So in normal configuration the current regulation will be greater for supply voltage and load variations.
$endgroup$
– WhatRoughBeast
15 hours ago
2
$begingroup$
It is good practice to number the components R1, R2, Q1, Q2, etc. so we can discuss them without confusion.
$endgroup$
– Transistor
15 hours ago
$begingroup$
@Transistor You are right, but the tool didn't allow me to add the component name. Apologies.
$endgroup$
– Embedded Geek
15 hours ago
|
show 8 more comments
2
$begingroup$
The polarity of your left-hand transistor is wrong. The transistor is operating in reverse mode (which it will do, just not very well). Try connecting the emitter to ground.
$endgroup$
– WhatRoughBeast
15 hours ago
$begingroup$
Right. What difference will it make by the way?
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
The transistor will have less gain in the reversed configuration than in the normal. So in normal configuration the current regulation will be greater for supply voltage and load variations.
$endgroup$
– WhatRoughBeast
15 hours ago
2
$begingroup$
It is good practice to number the components R1, R2, Q1, Q2, etc. so we can discuss them without confusion.
$endgroup$
– Transistor
15 hours ago
$begingroup$
@Transistor You are right, but the tool didn't allow me to add the component name. Apologies.
$endgroup$
– Embedded Geek
15 hours ago
2
2
$begingroup$
The polarity of your left-hand transistor is wrong. The transistor is operating in reverse mode (which it will do, just not very well). Try connecting the emitter to ground.
$endgroup$
– WhatRoughBeast
15 hours ago
$begingroup$
The polarity of your left-hand transistor is wrong. The transistor is operating in reverse mode (which it will do, just not very well). Try connecting the emitter to ground.
$endgroup$
– WhatRoughBeast
15 hours ago
$begingroup$
Right. What difference will it make by the way?
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
Right. What difference will it make by the way?
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
The transistor will have less gain in the reversed configuration than in the normal. So in normal configuration the current regulation will be greater for supply voltage and load variations.
$endgroup$
– WhatRoughBeast
15 hours ago
$begingroup$
The transistor will have less gain in the reversed configuration than in the normal. So in normal configuration the current regulation will be greater for supply voltage and load variations.
$endgroup$
– WhatRoughBeast
15 hours ago
2
2
$begingroup$
It is good practice to number the components R1, R2, Q1, Q2, etc. so we can discuss them without confusion.
$endgroup$
– Transistor
15 hours ago
$begingroup$
It is good practice to number the components R1, R2, Q1, Q2, etc. so we can discuss them without confusion.
$endgroup$
– Transistor
15 hours ago
$begingroup$
@Transistor You are right, but the tool didn't allow me to add the component name. Apologies.
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
@Transistor You are right, but the tool didn't allow me to add the component name. Apologies.
$endgroup$
– Embedded Geek
15 hours ago
|
show 8 more comments
5 Answers
5
active
oldest
votes
$begingroup$
Yes, the NPN transistor will heat up by exactly the same amount as a variable resistor if it were used to set the same LED current.
The circuit shown should work reasonably well down to a few volts but don’t just trust me on this - get a free sim tool and prove it yourself.
answered 15 hours ago
Andy akaAndy aka
243k11184418
$endgroup$
$begingroup$
Which free sim tool? Unable to find, any links?
$endgroup$
– Embedded Geek
15 hours ago
2
$begingroup$
@EmbeddedGeek circuitjs is popular and easy, or you could use the more powerful LTspice.
$endgroup$
– Hearth
15 hours ago
1
$begingroup$
or use falstad online simulator
$endgroup$
– Mitu Raj
15 hours ago
1
$begingroup$
@MituRaj the Falstad simulator is circuitjs.
$endgroup$
– Hearth
15 hours ago
2
$begingroup$
Nitpicking, but not exactly the same amount: it will get 0.5-1% extra heat from base current, and the 50 ohm resistor will reduce the heat a bit.
$endgroup$
– jpa
9 hours ago
|
show 2 more comments
$begingroup$
The circut will work and the main transistor will dissipate around 300mW and have a temperature of 55°C.
P = (Vc-Ve)*Ie
300mW = (17V-0.8V)*0.018mA
(You could use Ic, the diffrence will be very small since Ie=Ic+Ib and base current is insignificant.)
You should use a transistor able to dissipate more than 0.6W (ideally 1W), such as 2N3904.
The temperature can be calculated this way:
Tj=P*Rthj-amb (thermal resistance)
For 2N3904:
55°C=0.3W*200°C/W
This is hot to touch but will not damage the transistor, I suggest to put it away from the plastic case and capacitors.
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
$endgroup$
1
$begingroup$
FYI: 55 degrees C is not simply “hot to touch” it’s at the point of causing first degree burns in a few seconds. Not extremely hot for silicon, way too hot for skin.
$endgroup$
– Edgar Brown
12 hours ago
2
$begingroup$
Small typo. You wrote 0.018 mA instead of 18 mA.
$endgroup$
– Mitu Raj
10 hours ago
add a comment |
$begingroup$
Constant current for LED ...the voltage source to led will vary from 12V to 20V... the existing circuit is fine?
No, the circuit as shown DOES NOT WORK
I assume that the circuit should be as shown here:
simulate this circuit – Schematic created using CircuitLab
While you have not specified the devices used, I used a generic 2N3904 NPN here.
The 2N3904, Q1 would dissipate about 288mW and it has a maximum rating of 625mW.
At 25degC ambient you would expect the device chip to be at about 57degC and very hot to touch.
You could reduce the temperature of Q1 by adding a resistor for the LED, but all you are doing is spreading the dissipation between the resistor and Q1, you still have to dissipate 288mW.
A much better way to do this task is to use a single component such as the AL5809 constant current driver.
This is a single component, works from 2.5 - 60V and is cheap.
Update: It was suggested in comments that the circuit originally shown would work with the transistor in a reverse-active mode.
Maybe ….but the H(fe) would likely be in the single digit range (the 2N3904 is not exactly high gain) so trying to design the solution would be harder over the range of device parameters. See here.
Why anyone would suggest reverse active mode completely flummoxes me. There is no technical reason for choosing an unusual device operation mode for this type of simple application, so why even suggest it?
The feedback device (Q2 in my circuit) is never in Collector saturation, in addition it's never in Base/Emitter saturation, so there are no unusual configuration parameters.
The current does have a negative droop with temperature.
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
$endgroup$
$begingroup$
I understand I will be using, two resistors of 220 ohms each rated 1W in series with the LED, will dissipate much of the heat and much of the voltage, so Q1 won't be that much hot. Heat dissipation will spread between three components. Will that is fine?
$endgroup$
– Embedded Geek
12 hours ago
$begingroup$
I think it actually would work as drawn, with Q2 being in reverse-active mode... I almost wonder if someone drawing this circuit was trying to use the "the BE junction of a 3904 makes a decent zener diode" thing, since a zener diode between base and ground would also make sense there...
$endgroup$
– Hearth
12 hours ago
$begingroup$
@hearth Configured as 'Zener diode thing' makes no sense at all since the V(BE) junction is about a 6V Zener. How would you see this as working? ...It's clear that connecting an NPN in reverse mode would be equally futile, and would have absolutely no operational benefit.
$endgroup$
– Jack Creasey
12 hours ago
1
$begingroup$
@JackCreasey I don't think it makes sense either, I'm guessing as to what the person who drew the circuit may have been thinking. And with the transistor in reverse active mode, I just did some quick simulations and they seem to suggest that it still will work passably well as a current source. Of course, at that point, you're not getting much from having Q2 there and the circuit would work about equally well with it removed entirely, but it doesn't just stop being a current source.
$endgroup$
– Hearth
12 hours ago
$begingroup$
@EmbeddedGeek I'd use just a single 0.25W resistor in series with the LED. If you want the current to be constant between 12V and 20V then you don't want to drop any more than about 10V across the resistor. So a series resistor of 520 Ohms would seem appropriate.
$endgroup$
– Jack Creasey
12 hours ago
|
show 6 more comments
$begingroup$
Yes, it will work- it's a switched (not switching) crude linear regulator, but Q2 is operating in reverse mode and will regulate somewhat more poorly. It would be worse if your LED current was less since it has to sink about 3mA to throttle the base current of Q1, so that will consume about 1mA of base current in reverse active mode.
The transistor Q1 is dropping about 17.2V at 18.2mA so it will dissipate (ignoring base current dissipation, which is negligible) 313mW which is quite a bit for a TO-92 transistor. Assuming a 50°C maximum ambient and 200°C/W J-A thermal resistance, the junction will be at 112°C which is acceptable but not great for good reliability. If the '20V' was actually 30V it would be much worse, of course.
You can reduce the dissipation in the transistor by adding a resistor in series with the LED (based on the minimum the 20V supply could go such that you want to maintain regulation). For example, if it has to regulate from 15V to 20V you could add a resistor dropping about 5V, so 5/18.2 = 270$Omega$.
If the 20V isn't going to change much you hardly need a regulator at all, and just switching a resistor with a single transistor (which will then run cold) would be better and simpler (and more reliable). For example, a 1K 1/2-W resistor.
It's not a very good regulator because the "reference" is the Vbe of Q2 and it will vary with temperature. In some cases that could be considered a feature but I'll leave that discussion for another time.
Here is an LTspice simulation, however the temperature curves apply to the junction temperature of the transistor. As you can see, temperature is much more of an effect than the voltage (ignoring self-heating).
What will happen in fact is that there will be some self heating so the current will start off higher and tail off as Q1 heats Q2.
If Q2 is connected correctly, the LED current will be a bit lower because the base does not need so much current. My simulation (using the LTspice models) shows a bit lower current than yours so the dissipation is proportionally less. The major factor may be how optimistic their number is for reverse beta.
For a little post-prandial edit, I've added plots of Q1 dissipation in relation to the LED supply and temperature (blue lines).
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
$endgroup$
$begingroup$
Actually the temperature of Q1 will have little effect on the current, it's the Vf of the (if it's reversed-active mode) Q2 base-collector that dominates and provides the voltage reference. Q1 has almost nothing to do with the reference.
$endgroup$
– Jack Creasey
6 hours ago
$begingroup$
To show that Q2 V(BE) has no impact on the output current at all you could change Q2 for an N-channel FET and this would still regulate (though the reverse-active mode should no be used).
$endgroup$
– Jack Creasey
5 hours ago
$begingroup$
@JackCreasey That's why I was talking about the heat from Q1 reaching Q2 and thus changing the reference. Vgs(th) of a MOSFET is temperature sensitive too, and much more variable from unit to unit, so a BJT is the way to go.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
@mkeith Good question, I would guess it to be similar, though the Vf is higher because of the much higher base current so maybe it's worse, and maybe the resistive component and beta tempco cancel the diode part somewhat or maybe the add to it. The main difference I'm aware of is that Vce(sat) can be lower in reverse but that's of no consequence here since it never saturates. The current really is not all that repeatable because of the reverse beta so I would guess it's just an error.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
Yeah. I am hesitant to fully trust the simulation. But I am definitely not going to build the circuit to check it either. ;-)
$endgroup$
– mkeith
4 hours ago
|
show 2 more comments
$begingroup$
This is an actual constant current sink, based on a fixed base to ground voltage, which locks the Re voltage, so current per given Re is constant. The equation for current is $I = 0.65/R_e$.
simulate this circuit – Schematic created using CircuitLab
answered 8 hours ago
Sparky256Sparky256
12.2k21637
$endgroup$
$begingroup$
A poor configuration since the two diode thermal slopes are twice that of the transistor ….so as temp increases the current decreases markedly. Plus you'd need to feed R1 from the MCU if you wanted to be able to turn the CC driver off/on.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
@JackCreasey If D1 is thermally coupled to Q1, half this issue goes away. I have built many of these to drive LEDs at a constant brightness. I am not concerned about a fractional change in brightness.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
Trying to thermally couple a diode and transistor is almost pointless. The right way to do this is to use a shunt regulator or a CC diode.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
I could have added a cc source instead of a 2.2K resistor, but the OP just wants the basics, not a laboratory grade device.
$endgroup$
– Sparky256
8 hours ago
1
$begingroup$
Exactly, the simplest solution is one component....a CC regulator. two components if you want to turn it on/off. Rooting around for 49,9 Ohm resistors is so old hat.
$endgroup$
– Jack Creasey
7 hours ago
|
show 1 more comment
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, the NPN transistor will heat up by exactly the same amount as a variable resistor if it were used to set the same LED current.
The circuit shown should work reasonably well down to a few volts but don’t just trust me on this - get a free sim tool and prove it yourself.
answered 15 hours ago
Andy akaAndy aka
243k11184418
$endgroup$
$begingroup$
Which free sim tool? Unable to find, any links?
$endgroup$
– Embedded Geek
15 hours ago
2
$begingroup$
@EmbeddedGeek circuitjs is popular and easy, or you could use the more powerful LTspice.
$endgroup$
– Hearth
15 hours ago
1
$begingroup$
or use falstad online simulator
$endgroup$
– Mitu Raj
15 hours ago
1
$begingroup$
@MituRaj the Falstad simulator is circuitjs.
$endgroup$
– Hearth
15 hours ago
2
$begingroup$
Nitpicking, but not exactly the same amount: it will get 0.5-1% extra heat from base current, and the 50 ohm resistor will reduce the heat a bit.
$endgroup$
– jpa
9 hours ago
|
show 2 more comments
$begingroup$
Yes, the NPN transistor will heat up by exactly the same amount as a variable resistor if it were used to set the same LED current.
The circuit shown should work reasonably well down to a few volts but don’t just trust me on this - get a free sim tool and prove it yourself.
answered 15 hours ago
Andy akaAndy aka
243k11184418
$endgroup$
$begingroup$
Which free sim tool? Unable to find, any links?
$endgroup$
– Embedded Geek
15 hours ago
2
$begingroup$
@EmbeddedGeek circuitjs is popular and easy, or you could use the more powerful LTspice.
$endgroup$
– Hearth
15 hours ago
1
$begingroup$
or use falstad online simulator
$endgroup$
– Mitu Raj
15 hours ago
1
$begingroup$
@MituRaj the Falstad simulator is circuitjs.
$endgroup$
– Hearth
15 hours ago
2
$begingroup$
Nitpicking, but not exactly the same amount: it will get 0.5-1% extra heat from base current, and the 50 ohm resistor will reduce the heat a bit.
$endgroup$
– jpa
9 hours ago
|
show 2 more comments
$begingroup$
Yes, the NPN transistor will heat up by exactly the same amount as a variable resistor if it were used to set the same LED current.
The circuit shown should work reasonably well down to a few volts but don’t just trust me on this - get a free sim tool and prove it yourself.
answered 15 hours ago
Andy akaAndy aka
243k11184418
$endgroup$
Yes, the NPN transistor will heat up by exactly the same amount as a variable resistor if it were used to set the same LED current.
The circuit shown should work reasonably well down to a few volts but don’t just trust me on this - get a free sim tool and prove it yourself.
answered 15 hours ago
Andy akaAndy aka
243k11184418
answered 15 hours ago
Andy akaAndy aka
243k11184418
answered 15 hours ago
Andy akaAndy aka
243k11184418
answered 15 hours ago
Andy akaAndy aka
243k11184418
243k11184418
$begingroup$
Which free sim tool? Unable to find, any links?
$endgroup$
– Embedded Geek
15 hours ago
2
$begingroup$
@EmbeddedGeek circuitjs is popular and easy, or you could use the more powerful LTspice.
$endgroup$
– Hearth
15 hours ago
1
$begingroup$
or use falstad online simulator
$endgroup$
– Mitu Raj
15 hours ago
1
$begingroup$
@MituRaj the Falstad simulator is circuitjs.
$endgroup$
– Hearth
15 hours ago
2
$begingroup$
Nitpicking, but not exactly the same amount: it will get 0.5-1% extra heat from base current, and the 50 ohm resistor will reduce the heat a bit.
$endgroup$
– jpa
9 hours ago
|
show 2 more comments
$begingroup$
Which free sim tool? Unable to find, any links?
$endgroup$
– Embedded Geek
15 hours ago
2
$begingroup$
@EmbeddedGeek circuitjs is popular and easy, or you could use the more powerful LTspice.
$endgroup$
– Hearth
15 hours ago
1
$begingroup$
or use falstad online simulator
$endgroup$
– Mitu Raj
15 hours ago
1
$begingroup$
@MituRaj the Falstad simulator is circuitjs.
$endgroup$
– Hearth
15 hours ago
2
$begingroup$
Nitpicking, but not exactly the same amount: it will get 0.5-1% extra heat from base current, and the 50 ohm resistor will reduce the heat a bit.
$endgroup$
– jpa
9 hours ago
$begingroup$
Which free sim tool? Unable to find, any links?
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
Which free sim tool? Unable to find, any links?
$endgroup$
– Embedded Geek
15 hours ago
2
2
$begingroup$
@EmbeddedGeek circuitjs is popular and easy, or you could use the more powerful LTspice.
$endgroup$
– Hearth
15 hours ago
$begingroup$
@EmbeddedGeek circuitjs is popular and easy, or you could use the more powerful LTspice.
$endgroup$
– Hearth
15 hours ago
1
1
$begingroup$
or use falstad online simulator
$endgroup$
– Mitu Raj
15 hours ago
$begingroup$
or use falstad online simulator
$endgroup$
– Mitu Raj
15 hours ago
1
1
$begingroup$
@MituRaj the Falstad simulator is circuitjs.
$endgroup$
– Hearth
15 hours ago
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@MituRaj the Falstad simulator is circuitjs.
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– Hearth
15 hours ago
2
2
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Nitpicking, but not exactly the same amount: it will get 0.5-1% extra heat from base current, and the 50 ohm resistor will reduce the heat a bit.
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– jpa
9 hours ago
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Nitpicking, but not exactly the same amount: it will get 0.5-1% extra heat from base current, and the 50 ohm resistor will reduce the heat a bit.
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– jpa
9 hours ago
|
show 2 more comments
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The circut will work and the main transistor will dissipate around 300mW and have a temperature of 55°C.
P = (Vc-Ve)*Ie
300mW = (17V-0.8V)*0.018mA
(You could use Ic, the diffrence will be very small since Ie=Ic+Ib and base current is insignificant.)
You should use a transistor able to dissipate more than 0.6W (ideally 1W), such as 2N3904.
The temperature can be calculated this way:
Tj=P*Rthj-amb (thermal resistance)
For 2N3904:
55°C=0.3W*200°C/W
This is hot to touch but will not damage the transistor, I suggest to put it away from the plastic case and capacitors.
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
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1
$begingroup$
FYI: 55 degrees C is not simply “hot to touch” it’s at the point of causing first degree burns in a few seconds. Not extremely hot for silicon, way too hot for skin.
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– Edgar Brown
12 hours ago
2
$begingroup$
Small typo. You wrote 0.018 mA instead of 18 mA.
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– Mitu Raj
10 hours ago
add a comment |
$begingroup$
The circut will work and the main transistor will dissipate around 300mW and have a temperature of 55°C.
P = (Vc-Ve)*Ie
300mW = (17V-0.8V)*0.018mA
(You could use Ic, the diffrence will be very small since Ie=Ic+Ib and base current is insignificant.)
You should use a transistor able to dissipate more than 0.6W (ideally 1W), such as 2N3904.
The temperature can be calculated this way:
Tj=P*Rthj-amb (thermal resistance)
For 2N3904:
55°C=0.3W*200°C/W
This is hot to touch but will not damage the transistor, I suggest to put it away from the plastic case and capacitors.
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
$endgroup$
1
$begingroup$
FYI: 55 degrees C is not simply “hot to touch” it’s at the point of causing first degree burns in a few seconds. Not extremely hot for silicon, way too hot for skin.
$endgroup$
– Edgar Brown
12 hours ago
2
$begingroup$
Small typo. You wrote 0.018 mA instead of 18 mA.
$endgroup$
– Mitu Raj
10 hours ago
add a comment |
$begingroup$
The circut will work and the main transistor will dissipate around 300mW and have a temperature of 55°C.
P = (Vc-Ve)*Ie
300mW = (17V-0.8V)*0.018mA
(You could use Ic, the diffrence will be very small since Ie=Ic+Ib and base current is insignificant.)
You should use a transistor able to dissipate more than 0.6W (ideally 1W), such as 2N3904.
The temperature can be calculated this way:
Tj=P*Rthj-amb (thermal resistance)
For 2N3904:
55°C=0.3W*200°C/W
This is hot to touch but will not damage the transistor, I suggest to put it away from the plastic case and capacitors.
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
$endgroup$
The circut will work and the main transistor will dissipate around 300mW and have a temperature of 55°C.
P = (Vc-Ve)*Ie
300mW = (17V-0.8V)*0.018mA
(You could use Ic, the diffrence will be very small since Ie=Ic+Ib and base current is insignificant.)
You should use a transistor able to dissipate more than 0.6W (ideally 1W), such as 2N3904.
The temperature can be calculated this way:
Tj=P*Rthj-amb (thermal resistance)
For 2N3904:
55°C=0.3W*200°C/W
This is hot to touch but will not damage the transistor, I suggest to put it away from the plastic case and capacitors.
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
answered 13 hours ago
thegamebusterplthegamebusterpl
1188
1188
1
$begingroup$
FYI: 55 degrees C is not simply “hot to touch” it’s at the point of causing first degree burns in a few seconds. Not extremely hot for silicon, way too hot for skin.
$endgroup$
– Edgar Brown
12 hours ago
2
$begingroup$
Small typo. You wrote 0.018 mA instead of 18 mA.
$endgroup$
– Mitu Raj
10 hours ago
add a comment |
1
$begingroup$
FYI: 55 degrees C is not simply “hot to touch” it’s at the point of causing first degree burns in a few seconds. Not extremely hot for silicon, way too hot for skin.
$endgroup$
– Edgar Brown
12 hours ago
2
$begingroup$
Small typo. You wrote 0.018 mA instead of 18 mA.
$endgroup$
– Mitu Raj
10 hours ago
1
1
$begingroup$
FYI: 55 degrees C is not simply “hot to touch” it’s at the point of causing first degree burns in a few seconds. Not extremely hot for silicon, way too hot for skin.
$endgroup$
– Edgar Brown
12 hours ago
$begingroup$
FYI: 55 degrees C is not simply “hot to touch” it’s at the point of causing first degree burns in a few seconds. Not extremely hot for silicon, way too hot for skin.
$endgroup$
– Edgar Brown
12 hours ago
2
2
$begingroup$
Small typo. You wrote 0.018 mA instead of 18 mA.
$endgroup$
– Mitu Raj
10 hours ago
$begingroup$
Small typo. You wrote 0.018 mA instead of 18 mA.
$endgroup$
– Mitu Raj
10 hours ago
add a comment |
$begingroup$
Constant current for LED ...the voltage source to led will vary from 12V to 20V... the existing circuit is fine?
No, the circuit as shown DOES NOT WORK
I assume that the circuit should be as shown here:
simulate this circuit – Schematic created using CircuitLab
While you have not specified the devices used, I used a generic 2N3904 NPN here.
The 2N3904, Q1 would dissipate about 288mW and it has a maximum rating of 625mW.
At 25degC ambient you would expect the device chip to be at about 57degC and very hot to touch.
You could reduce the temperature of Q1 by adding a resistor for the LED, but all you are doing is spreading the dissipation between the resistor and Q1, you still have to dissipate 288mW.
A much better way to do this task is to use a single component such as the AL5809 constant current driver.
This is a single component, works from 2.5 - 60V and is cheap.
Update: It was suggested in comments that the circuit originally shown would work with the transistor in a reverse-active mode.
Maybe ….but the H(fe) would likely be in the single digit range (the 2N3904 is not exactly high gain) so trying to design the solution would be harder over the range of device parameters. See here.
Why anyone would suggest reverse active mode completely flummoxes me. There is no technical reason for choosing an unusual device operation mode for this type of simple application, so why even suggest it?
The feedback device (Q2 in my circuit) is never in Collector saturation, in addition it's never in Base/Emitter saturation, so there are no unusual configuration parameters.
The current does have a negative droop with temperature.
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
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I understand I will be using, two resistors of 220 ohms each rated 1W in series with the LED, will dissipate much of the heat and much of the voltage, so Q1 won't be that much hot. Heat dissipation will spread between three components. Will that is fine?
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– Embedded Geek
12 hours ago
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I think it actually would work as drawn, with Q2 being in reverse-active mode... I almost wonder if someone drawing this circuit was trying to use the "the BE junction of a 3904 makes a decent zener diode" thing, since a zener diode between base and ground would also make sense there...
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– Hearth
12 hours ago
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@hearth Configured as 'Zener diode thing' makes no sense at all since the V(BE) junction is about a 6V Zener. How would you see this as working? ...It's clear that connecting an NPN in reverse mode would be equally futile, and would have absolutely no operational benefit.
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– Jack Creasey
12 hours ago
1
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@JackCreasey I don't think it makes sense either, I'm guessing as to what the person who drew the circuit may have been thinking. And with the transistor in reverse active mode, I just did some quick simulations and they seem to suggest that it still will work passably well as a current source. Of course, at that point, you're not getting much from having Q2 there and the circuit would work about equally well with it removed entirely, but it doesn't just stop being a current source.
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– Hearth
12 hours ago
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@EmbeddedGeek I'd use just a single 0.25W resistor in series with the LED. If you want the current to be constant between 12V and 20V then you don't want to drop any more than about 10V across the resistor. So a series resistor of 520 Ohms would seem appropriate.
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– Jack Creasey
12 hours ago
|
show 6 more comments
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Constant current for LED ...the voltage source to led will vary from 12V to 20V... the existing circuit is fine?
No, the circuit as shown DOES NOT WORK
I assume that the circuit should be as shown here:
simulate this circuit – Schematic created using CircuitLab
While you have not specified the devices used, I used a generic 2N3904 NPN here.
The 2N3904, Q1 would dissipate about 288mW and it has a maximum rating of 625mW.
At 25degC ambient you would expect the device chip to be at about 57degC and very hot to touch.
You could reduce the temperature of Q1 by adding a resistor for the LED, but all you are doing is spreading the dissipation between the resistor and Q1, you still have to dissipate 288mW.
A much better way to do this task is to use a single component such as the AL5809 constant current driver.
This is a single component, works from 2.5 - 60V and is cheap.
Update: It was suggested in comments that the circuit originally shown would work with the transistor in a reverse-active mode.
Maybe ….but the H(fe) would likely be in the single digit range (the 2N3904 is not exactly high gain) so trying to design the solution would be harder over the range of device parameters. See here.
Why anyone would suggest reverse active mode completely flummoxes me. There is no technical reason for choosing an unusual device operation mode for this type of simple application, so why even suggest it?
The feedback device (Q2 in my circuit) is never in Collector saturation, in addition it's never in Base/Emitter saturation, so there are no unusual configuration parameters.
The current does have a negative droop with temperature.
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
$endgroup$
$begingroup$
I understand I will be using, two resistors of 220 ohms each rated 1W in series with the LED, will dissipate much of the heat and much of the voltage, so Q1 won't be that much hot. Heat dissipation will spread between three components. Will that is fine?
$endgroup$
– Embedded Geek
12 hours ago
$begingroup$
I think it actually would work as drawn, with Q2 being in reverse-active mode... I almost wonder if someone drawing this circuit was trying to use the "the BE junction of a 3904 makes a decent zener diode" thing, since a zener diode between base and ground would also make sense there...
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– Hearth
12 hours ago
$begingroup$
@hearth Configured as 'Zener diode thing' makes no sense at all since the V(BE) junction is about a 6V Zener. How would you see this as working? ...It's clear that connecting an NPN in reverse mode would be equally futile, and would have absolutely no operational benefit.
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– Jack Creasey
12 hours ago
1
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@JackCreasey I don't think it makes sense either, I'm guessing as to what the person who drew the circuit may have been thinking. And with the transistor in reverse active mode, I just did some quick simulations and they seem to suggest that it still will work passably well as a current source. Of course, at that point, you're not getting much from having Q2 there and the circuit would work about equally well with it removed entirely, but it doesn't just stop being a current source.
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– Hearth
12 hours ago
$begingroup$
@EmbeddedGeek I'd use just a single 0.25W resistor in series with the LED. If you want the current to be constant between 12V and 20V then you don't want to drop any more than about 10V across the resistor. So a series resistor of 520 Ohms would seem appropriate.
$endgroup$
– Jack Creasey
12 hours ago
|
show 6 more comments
$begingroup$
Constant current for LED ...the voltage source to led will vary from 12V to 20V... the existing circuit is fine?
No, the circuit as shown DOES NOT WORK
I assume that the circuit should be as shown here:
simulate this circuit – Schematic created using CircuitLab
While you have not specified the devices used, I used a generic 2N3904 NPN here.
The 2N3904, Q1 would dissipate about 288mW and it has a maximum rating of 625mW.
At 25degC ambient you would expect the device chip to be at about 57degC and very hot to touch.
You could reduce the temperature of Q1 by adding a resistor for the LED, but all you are doing is spreading the dissipation between the resistor and Q1, you still have to dissipate 288mW.
A much better way to do this task is to use a single component such as the AL5809 constant current driver.
This is a single component, works from 2.5 - 60V and is cheap.
Update: It was suggested in comments that the circuit originally shown would work with the transistor in a reverse-active mode.
Maybe ….but the H(fe) would likely be in the single digit range (the 2N3904 is not exactly high gain) so trying to design the solution would be harder over the range of device parameters. See here.
Why anyone would suggest reverse active mode completely flummoxes me. There is no technical reason for choosing an unusual device operation mode for this type of simple application, so why even suggest it?
The feedback device (Q2 in my circuit) is never in Collector saturation, in addition it's never in Base/Emitter saturation, so there are no unusual configuration parameters.
The current does have a negative droop with temperature.
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
$endgroup$
Constant current for LED ...the voltage source to led will vary from 12V to 20V... the existing circuit is fine?
No, the circuit as shown DOES NOT WORK
I assume that the circuit should be as shown here:
simulate this circuit – Schematic created using CircuitLab
While you have not specified the devices used, I used a generic 2N3904 NPN here.
The 2N3904, Q1 would dissipate about 288mW and it has a maximum rating of 625mW.
At 25degC ambient you would expect the device chip to be at about 57degC and very hot to touch.
You could reduce the temperature of Q1 by adding a resistor for the LED, but all you are doing is spreading the dissipation between the resistor and Q1, you still have to dissipate 288mW.
A much better way to do this task is to use a single component such as the AL5809 constant current driver.
This is a single component, works from 2.5 - 60V and is cheap.
Update: It was suggested in comments that the circuit originally shown would work with the transistor in a reverse-active mode.
Maybe ….but the H(fe) would likely be in the single digit range (the 2N3904 is not exactly high gain) so trying to design the solution would be harder over the range of device parameters. See here.
Why anyone would suggest reverse active mode completely flummoxes me. There is no technical reason for choosing an unusual device operation mode for this type of simple application, so why even suggest it?
The feedback device (Q2 in my circuit) is never in Collector saturation, in addition it's never in Base/Emitter saturation, so there are no unusual configuration parameters.
The current does have a negative droop with temperature.
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
edited 7 hours ago
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
answered 13 hours ago
Jack CreaseyJack Creasey
14.8k2823
14.8k2823
$begingroup$
I understand I will be using, two resistors of 220 ohms each rated 1W in series with the LED, will dissipate much of the heat and much of the voltage, so Q1 won't be that much hot. Heat dissipation will spread between three components. Will that is fine?
$endgroup$
– Embedded Geek
12 hours ago
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I think it actually would work as drawn, with Q2 being in reverse-active mode... I almost wonder if someone drawing this circuit was trying to use the "the BE junction of a 3904 makes a decent zener diode" thing, since a zener diode between base and ground would also make sense there...
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– Hearth
12 hours ago
$begingroup$
@hearth Configured as 'Zener diode thing' makes no sense at all since the V(BE) junction is about a 6V Zener. How would you see this as working? ...It's clear that connecting an NPN in reverse mode would be equally futile, and would have absolutely no operational benefit.
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– Jack Creasey
12 hours ago
1
$begingroup$
@JackCreasey I don't think it makes sense either, I'm guessing as to what the person who drew the circuit may have been thinking. And with the transistor in reverse active mode, I just did some quick simulations and they seem to suggest that it still will work passably well as a current source. Of course, at that point, you're not getting much from having Q2 there and the circuit would work about equally well with it removed entirely, but it doesn't just stop being a current source.
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– Hearth
12 hours ago
$begingroup$
@EmbeddedGeek I'd use just a single 0.25W resistor in series with the LED. If you want the current to be constant between 12V and 20V then you don't want to drop any more than about 10V across the resistor. So a series resistor of 520 Ohms would seem appropriate.
$endgroup$
– Jack Creasey
12 hours ago
|
show 6 more comments
$begingroup$
I understand I will be using, two resistors of 220 ohms each rated 1W in series with the LED, will dissipate much of the heat and much of the voltage, so Q1 won't be that much hot. Heat dissipation will spread between three components. Will that is fine?
$endgroup$
– Embedded Geek
12 hours ago
$begingroup$
I think it actually would work as drawn, with Q2 being in reverse-active mode... I almost wonder if someone drawing this circuit was trying to use the "the BE junction of a 3904 makes a decent zener diode" thing, since a zener diode between base and ground would also make sense there...
$endgroup$
– Hearth
12 hours ago
$begingroup$
@hearth Configured as 'Zener diode thing' makes no sense at all since the V(BE) junction is about a 6V Zener. How would you see this as working? ...It's clear that connecting an NPN in reverse mode would be equally futile, and would have absolutely no operational benefit.
$endgroup$
– Jack Creasey
12 hours ago
1
$begingroup$
@JackCreasey I don't think it makes sense either, I'm guessing as to what the person who drew the circuit may have been thinking. And with the transistor in reverse active mode, I just did some quick simulations and they seem to suggest that it still will work passably well as a current source. Of course, at that point, you're not getting much from having Q2 there and the circuit would work about equally well with it removed entirely, but it doesn't just stop being a current source.
$endgroup$
– Hearth
12 hours ago
$begingroup$
@EmbeddedGeek I'd use just a single 0.25W resistor in series with the LED. If you want the current to be constant between 12V and 20V then you don't want to drop any more than about 10V across the resistor. So a series resistor of 520 Ohms would seem appropriate.
$endgroup$
– Jack Creasey
12 hours ago
$begingroup$
I understand I will be using, two resistors of 220 ohms each rated 1W in series with the LED, will dissipate much of the heat and much of the voltage, so Q1 won't be that much hot. Heat dissipation will spread between three components. Will that is fine?
$endgroup$
– Embedded Geek
12 hours ago
$begingroup$
I understand I will be using, two resistors of 220 ohms each rated 1W in series with the LED, will dissipate much of the heat and much of the voltage, so Q1 won't be that much hot. Heat dissipation will spread between three components. Will that is fine?
$endgroup$
– Embedded Geek
12 hours ago
$begingroup$
I think it actually would work as drawn, with Q2 being in reverse-active mode... I almost wonder if someone drawing this circuit was trying to use the "the BE junction of a 3904 makes a decent zener diode" thing, since a zener diode between base and ground would also make sense there...
$endgroup$
– Hearth
12 hours ago
$begingroup$
I think it actually would work as drawn, with Q2 being in reverse-active mode... I almost wonder if someone drawing this circuit was trying to use the "the BE junction of a 3904 makes a decent zener diode" thing, since a zener diode between base and ground would also make sense there...
$endgroup$
– Hearth
12 hours ago
$begingroup$
@hearth Configured as 'Zener diode thing' makes no sense at all since the V(BE) junction is about a 6V Zener. How would you see this as working? ...It's clear that connecting an NPN in reverse mode would be equally futile, and would have absolutely no operational benefit.
$endgroup$
– Jack Creasey
12 hours ago
$begingroup$
@hearth Configured as 'Zener diode thing' makes no sense at all since the V(BE) junction is about a 6V Zener. How would you see this as working? ...It's clear that connecting an NPN in reverse mode would be equally futile, and would have absolutely no operational benefit.
$endgroup$
– Jack Creasey
12 hours ago
1
1
$begingroup$
@JackCreasey I don't think it makes sense either, I'm guessing as to what the person who drew the circuit may have been thinking. And with the transistor in reverse active mode, I just did some quick simulations and they seem to suggest that it still will work passably well as a current source. Of course, at that point, you're not getting much from having Q2 there and the circuit would work about equally well with it removed entirely, but it doesn't just stop being a current source.
$endgroup$
– Hearth
12 hours ago
$begingroup$
@JackCreasey I don't think it makes sense either, I'm guessing as to what the person who drew the circuit may have been thinking. And with the transistor in reverse active mode, I just did some quick simulations and they seem to suggest that it still will work passably well as a current source. Of course, at that point, you're not getting much from having Q2 there and the circuit would work about equally well with it removed entirely, but it doesn't just stop being a current source.
$endgroup$
– Hearth
12 hours ago
$begingroup$
@EmbeddedGeek I'd use just a single 0.25W resistor in series with the LED. If you want the current to be constant between 12V and 20V then you don't want to drop any more than about 10V across the resistor. So a series resistor of 520 Ohms would seem appropriate.
$endgroup$
– Jack Creasey
12 hours ago
$begingroup$
@EmbeddedGeek I'd use just a single 0.25W resistor in series with the LED. If you want the current to be constant between 12V and 20V then you don't want to drop any more than about 10V across the resistor. So a series resistor of 520 Ohms would seem appropriate.
$endgroup$
– Jack Creasey
12 hours ago
|
show 6 more comments
$begingroup$
Yes, it will work- it's a switched (not switching) crude linear regulator, but Q2 is operating in reverse mode and will regulate somewhat more poorly. It would be worse if your LED current was less since it has to sink about 3mA to throttle the base current of Q1, so that will consume about 1mA of base current in reverse active mode.
The transistor Q1 is dropping about 17.2V at 18.2mA so it will dissipate (ignoring base current dissipation, which is negligible) 313mW which is quite a bit for a TO-92 transistor. Assuming a 50°C maximum ambient and 200°C/W J-A thermal resistance, the junction will be at 112°C which is acceptable but not great for good reliability. If the '20V' was actually 30V it would be much worse, of course.
You can reduce the dissipation in the transistor by adding a resistor in series with the LED (based on the minimum the 20V supply could go such that you want to maintain regulation). For example, if it has to regulate from 15V to 20V you could add a resistor dropping about 5V, so 5/18.2 = 270$Omega$.
If the 20V isn't going to change much you hardly need a regulator at all, and just switching a resistor with a single transistor (which will then run cold) would be better and simpler (and more reliable). For example, a 1K 1/2-W resistor.
It's not a very good regulator because the "reference" is the Vbe of Q2 and it will vary with temperature. In some cases that could be considered a feature but I'll leave that discussion for another time.
Here is an LTspice simulation, however the temperature curves apply to the junction temperature of the transistor. As you can see, temperature is much more of an effect than the voltage (ignoring self-heating).
What will happen in fact is that there will be some self heating so the current will start off higher and tail off as Q1 heats Q2.
If Q2 is connected correctly, the LED current will be a bit lower because the base does not need so much current. My simulation (using the LTspice models) shows a bit lower current than yours so the dissipation is proportionally less. The major factor may be how optimistic their number is for reverse beta.
For a little post-prandial edit, I've added plots of Q1 dissipation in relation to the LED supply and temperature (blue lines).
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
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Actually the temperature of Q1 will have little effect on the current, it's the Vf of the (if it's reversed-active mode) Q2 base-collector that dominates and provides the voltage reference. Q1 has almost nothing to do with the reference.
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– Jack Creasey
6 hours ago
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To show that Q2 V(BE) has no impact on the output current at all you could change Q2 for an N-channel FET and this would still regulate (though the reverse-active mode should no be used).
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– Jack Creasey
5 hours ago
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@JackCreasey That's why I was talking about the heat from Q1 reaching Q2 and thus changing the reference. Vgs(th) of a MOSFET is temperature sensitive too, and much more variable from unit to unit, so a BJT is the way to go.
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– Spehro Pefhany
5 hours ago
1
$begingroup$
@mkeith Good question, I would guess it to be similar, though the Vf is higher because of the much higher base current so maybe it's worse, and maybe the resistive component and beta tempco cancel the diode part somewhat or maybe the add to it. The main difference I'm aware of is that Vce(sat) can be lower in reverse but that's of no consequence here since it never saturates. The current really is not all that repeatable because of the reverse beta so I would guess it's just an error.
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– Spehro Pefhany
5 hours ago
1
$begingroup$
Yeah. I am hesitant to fully trust the simulation. But I am definitely not going to build the circuit to check it either. ;-)
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– mkeith
4 hours ago
|
show 2 more comments
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Yes, it will work- it's a switched (not switching) crude linear regulator, but Q2 is operating in reverse mode and will regulate somewhat more poorly. It would be worse if your LED current was less since it has to sink about 3mA to throttle the base current of Q1, so that will consume about 1mA of base current in reverse active mode.
The transistor Q1 is dropping about 17.2V at 18.2mA so it will dissipate (ignoring base current dissipation, which is negligible) 313mW which is quite a bit for a TO-92 transistor. Assuming a 50°C maximum ambient and 200°C/W J-A thermal resistance, the junction will be at 112°C which is acceptable but not great for good reliability. If the '20V' was actually 30V it would be much worse, of course.
You can reduce the dissipation in the transistor by adding a resistor in series with the LED (based on the minimum the 20V supply could go such that you want to maintain regulation). For example, if it has to regulate from 15V to 20V you could add a resistor dropping about 5V, so 5/18.2 = 270$Omega$.
If the 20V isn't going to change much you hardly need a regulator at all, and just switching a resistor with a single transistor (which will then run cold) would be better and simpler (and more reliable). For example, a 1K 1/2-W resistor.
It's not a very good regulator because the "reference" is the Vbe of Q2 and it will vary with temperature. In some cases that could be considered a feature but I'll leave that discussion for another time.
Here is an LTspice simulation, however the temperature curves apply to the junction temperature of the transistor. As you can see, temperature is much more of an effect than the voltage (ignoring self-heating).
What will happen in fact is that there will be some self heating so the current will start off higher and tail off as Q1 heats Q2.
If Q2 is connected correctly, the LED current will be a bit lower because the base does not need so much current. My simulation (using the LTspice models) shows a bit lower current than yours so the dissipation is proportionally less. The major factor may be how optimistic their number is for reverse beta.
For a little post-prandial edit, I've added plots of Q1 dissipation in relation to the LED supply and temperature (blue lines).
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
$endgroup$
$begingroup$
Actually the temperature of Q1 will have little effect on the current, it's the Vf of the (if it's reversed-active mode) Q2 base-collector that dominates and provides the voltage reference. Q1 has almost nothing to do with the reference.
$endgroup$
– Jack Creasey
6 hours ago
$begingroup$
To show that Q2 V(BE) has no impact on the output current at all you could change Q2 for an N-channel FET and this would still regulate (though the reverse-active mode should no be used).
$endgroup$
– Jack Creasey
5 hours ago
$begingroup$
@JackCreasey That's why I was talking about the heat from Q1 reaching Q2 and thus changing the reference. Vgs(th) of a MOSFET is temperature sensitive too, and much more variable from unit to unit, so a BJT is the way to go.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
@mkeith Good question, I would guess it to be similar, though the Vf is higher because of the much higher base current so maybe it's worse, and maybe the resistive component and beta tempco cancel the diode part somewhat or maybe the add to it. The main difference I'm aware of is that Vce(sat) can be lower in reverse but that's of no consequence here since it never saturates. The current really is not all that repeatable because of the reverse beta so I would guess it's just an error.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
Yeah. I am hesitant to fully trust the simulation. But I am definitely not going to build the circuit to check it either. ;-)
$endgroup$
– mkeith
4 hours ago
|
show 2 more comments
$begingroup$
Yes, it will work- it's a switched (not switching) crude linear regulator, but Q2 is operating in reverse mode and will regulate somewhat more poorly. It would be worse if your LED current was less since it has to sink about 3mA to throttle the base current of Q1, so that will consume about 1mA of base current in reverse active mode.
The transistor Q1 is dropping about 17.2V at 18.2mA so it will dissipate (ignoring base current dissipation, which is negligible) 313mW which is quite a bit for a TO-92 transistor. Assuming a 50°C maximum ambient and 200°C/W J-A thermal resistance, the junction will be at 112°C which is acceptable but not great for good reliability. If the '20V' was actually 30V it would be much worse, of course.
You can reduce the dissipation in the transistor by adding a resistor in series with the LED (based on the minimum the 20V supply could go such that you want to maintain regulation). For example, if it has to regulate from 15V to 20V you could add a resistor dropping about 5V, so 5/18.2 = 270$Omega$.
If the 20V isn't going to change much you hardly need a regulator at all, and just switching a resistor with a single transistor (which will then run cold) would be better and simpler (and more reliable). For example, a 1K 1/2-W resistor.
It's not a very good regulator because the "reference" is the Vbe of Q2 and it will vary with temperature. In some cases that could be considered a feature but I'll leave that discussion for another time.
Here is an LTspice simulation, however the temperature curves apply to the junction temperature of the transistor. As you can see, temperature is much more of an effect than the voltage (ignoring self-heating).
What will happen in fact is that there will be some self heating so the current will start off higher and tail off as Q1 heats Q2.
If Q2 is connected correctly, the LED current will be a bit lower because the base does not need so much current. My simulation (using the LTspice models) shows a bit lower current than yours so the dissipation is proportionally less. The major factor may be how optimistic their number is for reverse beta.
For a little post-prandial edit, I've added plots of Q1 dissipation in relation to the LED supply and temperature (blue lines).
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
$endgroup$
Yes, it will work- it's a switched (not switching) crude linear regulator, but Q2 is operating in reverse mode and will regulate somewhat more poorly. It would be worse if your LED current was less since it has to sink about 3mA to throttle the base current of Q1, so that will consume about 1mA of base current in reverse active mode.
The transistor Q1 is dropping about 17.2V at 18.2mA so it will dissipate (ignoring base current dissipation, which is negligible) 313mW which is quite a bit for a TO-92 transistor. Assuming a 50°C maximum ambient and 200°C/W J-A thermal resistance, the junction will be at 112°C which is acceptable but not great for good reliability. If the '20V' was actually 30V it would be much worse, of course.
You can reduce the dissipation in the transistor by adding a resistor in series with the LED (based on the minimum the 20V supply could go such that you want to maintain regulation). For example, if it has to regulate from 15V to 20V you could add a resistor dropping about 5V, so 5/18.2 = 270$Omega$.
If the 20V isn't going to change much you hardly need a regulator at all, and just switching a resistor with a single transistor (which will then run cold) would be better and simpler (and more reliable). For example, a 1K 1/2-W resistor.
It's not a very good regulator because the "reference" is the Vbe of Q2 and it will vary with temperature. In some cases that could be considered a feature but I'll leave that discussion for another time.
Here is an LTspice simulation, however the temperature curves apply to the junction temperature of the transistor. As you can see, temperature is much more of an effect than the voltage (ignoring self-heating).
What will happen in fact is that there will be some self heating so the current will start off higher and tail off as Q1 heats Q2.
If Q2 is connected correctly, the LED current will be a bit lower because the base does not need so much current. My simulation (using the LTspice models) shows a bit lower current than yours so the dissipation is proportionally less. The major factor may be how optimistic their number is for reverse beta.
For a little post-prandial edit, I've added plots of Q1 dissipation in relation to the LED supply and temperature (blue lines).
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
edited 5 hours ago
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
answered 6 hours ago
Spehro PefhanySpehro Pefhany
210k5162425
210k5162425
$begingroup$
Actually the temperature of Q1 will have little effect on the current, it's the Vf of the (if it's reversed-active mode) Q2 base-collector that dominates and provides the voltage reference. Q1 has almost nothing to do with the reference.
$endgroup$
– Jack Creasey
6 hours ago
$begingroup$
To show that Q2 V(BE) has no impact on the output current at all you could change Q2 for an N-channel FET and this would still regulate (though the reverse-active mode should no be used).
$endgroup$
– Jack Creasey
5 hours ago
$begingroup$
@JackCreasey That's why I was talking about the heat from Q1 reaching Q2 and thus changing the reference. Vgs(th) of a MOSFET is temperature sensitive too, and much more variable from unit to unit, so a BJT is the way to go.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
@mkeith Good question, I would guess it to be similar, though the Vf is higher because of the much higher base current so maybe it's worse, and maybe the resistive component and beta tempco cancel the diode part somewhat or maybe the add to it. The main difference I'm aware of is that Vce(sat) can be lower in reverse but that's of no consequence here since it never saturates. The current really is not all that repeatable because of the reverse beta so I would guess it's just an error.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
Yeah. I am hesitant to fully trust the simulation. But I am definitely not going to build the circuit to check it either. ;-)
$endgroup$
– mkeith
4 hours ago
|
show 2 more comments
$begingroup$
Actually the temperature of Q1 will have little effect on the current, it's the Vf of the (if it's reversed-active mode) Q2 base-collector that dominates and provides the voltage reference. Q1 has almost nothing to do with the reference.
$endgroup$
– Jack Creasey
6 hours ago
$begingroup$
To show that Q2 V(BE) has no impact on the output current at all you could change Q2 for an N-channel FET and this would still regulate (though the reverse-active mode should no be used).
$endgroup$
– Jack Creasey
5 hours ago
$begingroup$
@JackCreasey That's why I was talking about the heat from Q1 reaching Q2 and thus changing the reference. Vgs(th) of a MOSFET is temperature sensitive too, and much more variable from unit to unit, so a BJT is the way to go.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
@mkeith Good question, I would guess it to be similar, though the Vf is higher because of the much higher base current so maybe it's worse, and maybe the resistive component and beta tempco cancel the diode part somewhat or maybe the add to it. The main difference I'm aware of is that Vce(sat) can be lower in reverse but that's of no consequence here since it never saturates. The current really is not all that repeatable because of the reverse beta so I would guess it's just an error.
$endgroup$
– Spehro Pefhany
5 hours ago
1
$begingroup$
Yeah. I am hesitant to fully trust the simulation. But I am definitely not going to build the circuit to check it either. ;-)
$endgroup$
– mkeith
4 hours ago
$begingroup$
Actually the temperature of Q1 will have little effect on the current, it's the Vf of the (if it's reversed-active mode) Q2 base-collector that dominates and provides the voltage reference. Q1 has almost nothing to do with the reference.
$endgroup$
– Jack Creasey
6 hours ago
$begingroup$
Actually the temperature of Q1 will have little effect on the current, it's the Vf of the (if it's reversed-active mode) Q2 base-collector that dominates and provides the voltage reference. Q1 has almost nothing to do with the reference.
$endgroup$
– Jack Creasey
6 hours ago
$begingroup$
To show that Q2 V(BE) has no impact on the output current at all you could change Q2 for an N-channel FET and this would still regulate (though the reverse-active mode should no be used).
$endgroup$
– Jack Creasey
5 hours ago
$begingroup$
To show that Q2 V(BE) has no impact on the output current at all you could change Q2 for an N-channel FET and this would still regulate (though the reverse-active mode should no be used).
$endgroup$
– Jack Creasey
5 hours ago
$begingroup$
@JackCreasey That's why I was talking about the heat from Q1 reaching Q2 and thus changing the reference. Vgs(th) of a MOSFET is temperature sensitive too, and much more variable from unit to unit, so a BJT is the way to go.
$endgroup$
– Spehro Pefhany
5 hours ago
$begingroup$
@JackCreasey That's why I was talking about the heat from Q1 reaching Q2 and thus changing the reference. Vgs(th) of a MOSFET is temperature sensitive too, and much more variable from unit to unit, so a BJT is the way to go.
$endgroup$
– Spehro Pefhany
5 hours ago
1
1
$begingroup$
@mkeith Good question, I would guess it to be similar, though the Vf is higher because of the much higher base current so maybe it's worse, and maybe the resistive component and beta tempco cancel the diode part somewhat or maybe the add to it. The main difference I'm aware of is that Vce(sat) can be lower in reverse but that's of no consequence here since it never saturates. The current really is not all that repeatable because of the reverse beta so I would guess it's just an error.
$endgroup$
– Spehro Pefhany
5 hours ago
$begingroup$
@mkeith Good question, I would guess it to be similar, though the Vf is higher because of the much higher base current so maybe it's worse, and maybe the resistive component and beta tempco cancel the diode part somewhat or maybe the add to it. The main difference I'm aware of is that Vce(sat) can be lower in reverse but that's of no consequence here since it never saturates. The current really is not all that repeatable because of the reverse beta so I would guess it's just an error.
$endgroup$
– Spehro Pefhany
5 hours ago
1
1
$begingroup$
Yeah. I am hesitant to fully trust the simulation. But I am definitely not going to build the circuit to check it either. ;-)
$endgroup$
– mkeith
4 hours ago
$begingroup$
Yeah. I am hesitant to fully trust the simulation. But I am definitely not going to build the circuit to check it either. ;-)
$endgroup$
– mkeith
4 hours ago
|
show 2 more comments
$begingroup$
This is an actual constant current sink, based on a fixed base to ground voltage, which locks the Re voltage, so current per given Re is constant. The equation for current is $I = 0.65/R_e$.
simulate this circuit – Schematic created using CircuitLab
answered 8 hours ago
Sparky256Sparky256
12.2k21637
$endgroup$
$begingroup$
A poor configuration since the two diode thermal slopes are twice that of the transistor ….so as temp increases the current decreases markedly. Plus you'd need to feed R1 from the MCU if you wanted to be able to turn the CC driver off/on.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
@JackCreasey If D1 is thermally coupled to Q1, half this issue goes away. I have built many of these to drive LEDs at a constant brightness. I am not concerned about a fractional change in brightness.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
Trying to thermally couple a diode and transistor is almost pointless. The right way to do this is to use a shunt regulator or a CC diode.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
I could have added a cc source instead of a 2.2K resistor, but the OP just wants the basics, not a laboratory grade device.
$endgroup$
– Sparky256
8 hours ago
1
$begingroup$
Exactly, the simplest solution is one component....a CC regulator. two components if you want to turn it on/off. Rooting around for 49,9 Ohm resistors is so old hat.
$endgroup$
– Jack Creasey
7 hours ago
|
show 1 more comment
$begingroup$
This is an actual constant current sink, based on a fixed base to ground voltage, which locks the Re voltage, so current per given Re is constant. The equation for current is $I = 0.65/R_e$.
simulate this circuit – Schematic created using CircuitLab
answered 8 hours ago
Sparky256Sparky256
12.2k21637
$endgroup$
$begingroup$
A poor configuration since the two diode thermal slopes are twice that of the transistor ….so as temp increases the current decreases markedly. Plus you'd need to feed R1 from the MCU if you wanted to be able to turn the CC driver off/on.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
@JackCreasey If D1 is thermally coupled to Q1, half this issue goes away. I have built many of these to drive LEDs at a constant brightness. I am not concerned about a fractional change in brightness.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
Trying to thermally couple a diode and transistor is almost pointless. The right way to do this is to use a shunt regulator or a CC diode.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
I could have added a cc source instead of a 2.2K resistor, but the OP just wants the basics, not a laboratory grade device.
$endgroup$
– Sparky256
8 hours ago
1
$begingroup$
Exactly, the simplest solution is one component....a CC regulator. two components if you want to turn it on/off. Rooting around for 49,9 Ohm resistors is so old hat.
$endgroup$
– Jack Creasey
7 hours ago
|
show 1 more comment
$begingroup$
This is an actual constant current sink, based on a fixed base to ground voltage, which locks the Re voltage, so current per given Re is constant. The equation for current is $I = 0.65/R_e$.
simulate this circuit – Schematic created using CircuitLab
answered 8 hours ago
Sparky256Sparky256
12.2k21637
$endgroup$
This is an actual constant current sink, based on a fixed base to ground voltage, which locks the Re voltage, so current per given Re is constant. The equation for current is $I = 0.65/R_e$.
simulate this circuit – Schematic created using CircuitLab
answered 8 hours ago
Sparky256Sparky256
12.2k21637
edited 8 hours ago
answered 8 hours ago
Sparky256Sparky256
12.2k21637
answered 8 hours ago
Sparky256Sparky256
12.2k21637
answered 8 hours ago
Sparky256Sparky256
12.2k21637
12.2k21637
$begingroup$
A poor configuration since the two diode thermal slopes are twice that of the transistor ….so as temp increases the current decreases markedly. Plus you'd need to feed R1 from the MCU if you wanted to be able to turn the CC driver off/on.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
@JackCreasey If D1 is thermally coupled to Q1, half this issue goes away. I have built many of these to drive LEDs at a constant brightness. I am not concerned about a fractional change in brightness.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
Trying to thermally couple a diode and transistor is almost pointless. The right way to do this is to use a shunt regulator or a CC diode.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
I could have added a cc source instead of a 2.2K resistor, but the OP just wants the basics, not a laboratory grade device.
$endgroup$
– Sparky256
8 hours ago
1
$begingroup$
Exactly, the simplest solution is one component....a CC regulator. two components if you want to turn it on/off. Rooting around for 49,9 Ohm resistors is so old hat.
$endgroup$
– Jack Creasey
7 hours ago
|
show 1 more comment
$begingroup$
A poor configuration since the two diode thermal slopes are twice that of the transistor ….so as temp increases the current decreases markedly. Plus you'd need to feed R1 from the MCU if you wanted to be able to turn the CC driver off/on.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
@JackCreasey If D1 is thermally coupled to Q1, half this issue goes away. I have built many of these to drive LEDs at a constant brightness. I am not concerned about a fractional change in brightness.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
Trying to thermally couple a diode and transistor is almost pointless. The right way to do this is to use a shunt regulator or a CC diode.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
I could have added a cc source instead of a 2.2K resistor, but the OP just wants the basics, not a laboratory grade device.
$endgroup$
– Sparky256
8 hours ago
1
$begingroup$
Exactly, the simplest solution is one component....a CC regulator. two components if you want to turn it on/off. Rooting around for 49,9 Ohm resistors is so old hat.
$endgroup$
– Jack Creasey
7 hours ago
$begingroup$
A poor configuration since the two diode thermal slopes are twice that of the transistor ….so as temp increases the current decreases markedly. Plus you'd need to feed R1 from the MCU if you wanted to be able to turn the CC driver off/on.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
A poor configuration since the two diode thermal slopes are twice that of the transistor ….so as temp increases the current decreases markedly. Plus you'd need to feed R1 from the MCU if you wanted to be able to turn the CC driver off/on.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
@JackCreasey If D1 is thermally coupled to Q1, half this issue goes away. I have built many of these to drive LEDs at a constant brightness. I am not concerned about a fractional change in brightness.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
@JackCreasey If D1 is thermally coupled to Q1, half this issue goes away. I have built many of these to drive LEDs at a constant brightness. I am not concerned about a fractional change in brightness.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
Trying to thermally couple a diode and transistor is almost pointless. The right way to do this is to use a shunt regulator or a CC diode.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
Trying to thermally couple a diode and transistor is almost pointless. The right way to do this is to use a shunt regulator or a CC diode.
$endgroup$
– Jack Creasey
8 hours ago
$begingroup$
I could have added a cc source instead of a 2.2K resistor, but the OP just wants the basics, not a laboratory grade device.
$endgroup$
– Sparky256
8 hours ago
$begingroup$
I could have added a cc source instead of a 2.2K resistor, but the OP just wants the basics, not a laboratory grade device.
$endgroup$
– Sparky256
8 hours ago
1
1
$begingroup$
Exactly, the simplest solution is one component....a CC regulator. two components if you want to turn it on/off. Rooting around for 49,9 Ohm resistors is so old hat.
$endgroup$
– Jack Creasey
7 hours ago
$begingroup$
Exactly, the simplest solution is one component....a CC regulator. two components if you want to turn it on/off. Rooting around for 49,9 Ohm resistors is so old hat.
$endgroup$
– Jack Creasey
7 hours ago
|
show 1 more comment
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2
$begingroup$
The polarity of your left-hand transistor is wrong. The transistor is operating in reverse mode (which it will do, just not very well). Try connecting the emitter to ground.
$endgroup$
– WhatRoughBeast
15 hours ago
$begingroup$
Right. What difference will it make by the way?
$endgroup$
– Embedded Geek
15 hours ago
$begingroup$
The transistor will have less gain in the reversed configuration than in the normal. So in normal configuration the current regulation will be greater for supply voltage and load variations.
$endgroup$
– WhatRoughBeast
15 hours ago
2
$begingroup$
It is good practice to number the components R1, R2, Q1, Q2, etc. so we can discuss them without confusion.
$endgroup$
– Transistor
15 hours ago
$begingroup$
@Transistor You are right, but the tool didn't allow me to add the component name. Apologies.
$endgroup$
– Embedded Geek
15 hours ago